what is the specific heat of a 25.0 g substance that absorbs 493.4 j and raises the temperature from 12.0°c to 34°c?

Specific heat of substance  = 0.897 J/g.°C

Explanation:

Given data:

Specific heat of substance = ?

Mass of substance = 25.0 g

Heat absorbed = 493.4 J

Initial temperature = 12.0 °C

Final temperature = 34°C

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 34°C -12.0°C

ΔT = 22°C

493.4 J  = 25 g ×c× 22°C

493.4 J  = 550 g.°C ×c

c = 493.4 J  / 550 g.°C

c = 0.897 J/g.°C

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.